How to solve $2 = x \tan(x)$?

Question

When I plot the equation $2=x\tan x$ on Mathematica, I am seeing multiple zeros. But I am stumped as to how to find these zeros on paper.

Answer 1

$$f(x)= x\tan(x)-2$$

$$f'(x) = \tan(x) + x sec^2 (x) $$

Start at $x_0 =1$ and apply Newton Method, $$x_{n+1} = x_n - \frac {x_n \tan x_n -2}{\tan (x_n)+ \sec ^2 ( x_n)}$$

You will find one approximate solution by iteration.

$$1\to 1.088821754 \to 1.076426157\to 1.076898553\to 1.076872655.....1.076873986 \to 1.076873986$$ Other solutions are found with different initial point.

Answer 2

As said in comments, to find the zeros of function $$f(x)=x \tan(x)-2$$ you need some numerical methods. Since there are infinitely many, plot the function, locate where is more or less the root and use this value as a guess.

Since there is a symmetry, just focus on the positive ones.

In any manner, to have a better conditioning, it would be better to search for the zeros of function $$g(x)=x \sin(x)-2 \cos(x)$$ which does not show vertical asymptotes.

For example, there is a root close to $x=10$. Newton iterates would then be $$\left( \begin{array}{cc} n & x_n \\ 0 & 10.00000000 \\ 1 & 9.624648195 \\ 2 & 9.629562740 \\ 3 & 9.629560343 \end{array} \right)$$

Answer 3

We can also try to get to an approximate solution of all the roots, by finding an asymptotic expression for the large roots. It is easy to see that apart from the roots closest to the origin at approximately $x = \pm 1.076873986311803658608597673048696758216\cdots$, the other positive roots are at approximately:

$$x_n \approx \pi n$$

for $n$ an integer larger than zero.

We can then improve this estimate by performing a series expansion around $x = \pi n$. If we write the $n$th root as:

$$u_n \approx \pi n + a_0 + \frac{a_1}{n} + \frac{a_2}{n^2}+\cdots $$

and expand in powers of $\dfrac{1}{n}$, then we find that the coefficients $a_k$ for even $k$ are zero. The first few terms are as follows:

$$u_n =\pi n + \frac{2}{\pi n} -\frac{20}{3 \pi ^3 n^3} +\frac{656}{15 \pi ^5 n^5} -\frac{37424}{105 \pi ^7 n^7} +\frac{204352}{63 \pi ^9 n^9} -\frac{109411712}{3465 \pi ^{11} n^{11}}+\cdots$$